Problem: You have found the following ages (in years) of 5 zebras. Those zebras were randomly selected from the 32 zebras at your local zoo: $ 10,\enspace 30,\enspace 21,\enspace 13,\enspace 13$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 32 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{10 + 30 + 21 + 13 + 13}{{5}} = {17.4\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {54.76} + {158.76} + {12.96} + {19.36} + {19.36}} {{5 - 1}} $ {s^2} = \dfrac{{265.2}}{{4}} = {66.3\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{66.3\text{ years}^2}} = {8.1\text{ years}} $ We can estimate that the average zebra at the zoo is 17.4 years old. There is also a standard deviation of 8.1 years.